3.8 \(\int \frac {\coth ^{-1}(a x)}{x^2} \, dx\)

Optimal. Leaf size=30 \[ -\frac {1}{2} a \log \left (1-a^2 x^2\right )+a \log (x)-\frac {\coth ^{-1}(a x)}{x} \]

[Out]

-arccoth(a*x)/x+a*ln(x)-1/2*a*ln(-a^2*x^2+1)

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Rubi [A]  time = 0.02, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5917, 266, 36, 29, 31} \[ -\frac {1}{2} a \log \left (1-a^2 x^2\right )+a \log (x)-\frac {\coth ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[a*x]/x^2,x]

[Out]

-(ArcCoth[a*x]/x) + a*Log[x] - (a*Log[1 - a^2*x^2])/2

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5917

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcC
oth[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCoth[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\coth ^{-1}(a x)}{x^2} \, dx &=-\frac {\coth ^{-1}(a x)}{x}+a \int \frac {1}{x \left (1-a^2 x^2\right )} \, dx\\ &=-\frac {\coth ^{-1}(a x)}{x}+\frac {1}{2} a \operatorname {Subst}\left (\int \frac {1}{x \left (1-a^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {\coth ^{-1}(a x)}{x}+\frac {1}{2} a \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} a^3 \operatorname {Subst}\left (\int \frac {1}{1-a^2 x} \, dx,x,x^2\right )\\ &=-\frac {\coth ^{-1}(a x)}{x}+a \log (x)-\frac {1}{2} a \log \left (1-a^2 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 30, normalized size = 1.00 \[ -\frac {1}{2} a \log \left (1-a^2 x^2\right )+a \log (x)-\frac {\coth ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[a*x]/x^2,x]

[Out]

-(ArcCoth[a*x]/x) + a*Log[x] - (a*Log[1 - a^2*x^2])/2

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fricas [A]  time = 0.45, size = 39, normalized size = 1.30 \[ -\frac {a x \log \left (a^{2} x^{2} - 1\right ) - 2 \, a x \log \relax (x) + \log \left (\frac {a x + 1}{a x - 1}\right )}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)/x^2,x, algorithm="fricas")

[Out]

-1/2*(a*x*log(a^2*x^2 - 1) - 2*a*x*log(x) + log((a*x + 1)/(a*x - 1)))/x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcoth}\left (a x\right )}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)/x^2,x, algorithm="giac")

[Out]

integrate(arccoth(a*x)/x^2, x)

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maple [A]  time = 0.04, size = 35, normalized size = 1.17 \[ -\frac {\mathrm {arccoth}\left (a x \right )}{x}+a \ln \left (a x \right )-\frac {a \ln \left (a x -1\right )}{2}-\frac {a \ln \left (a x +1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(a*x)/x^2,x)

[Out]

-arccoth(a*x)/x+a*ln(a*x)-1/2*a*ln(a*x-1)-1/2*a*ln(a*x+1)

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maxima [A]  time = 0.31, size = 30, normalized size = 1.00 \[ -\frac {1}{2} \, a {\left (\log \left (a^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} - \frac {\operatorname {arcoth}\left (a x\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(a*x)/x^2,x, algorithm="maxima")

[Out]

-1/2*a*(log(a^2*x^2 - 1) - log(x^2)) - arccoth(a*x)/x

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mupad [B]  time = 1.16, size = 27, normalized size = 0.90 \[ a\,\ln \relax (x)-\frac {a\,\ln \left (a^2\,x^2-1\right )}{2}-\frac {\mathrm {acoth}\left (a\,x\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(a*x)/x^2,x)

[Out]

a*log(x) - (a*log(a^2*x^2 - 1))/2 - acoth(a*x)/x

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sympy [A]  time = 0.37, size = 26, normalized size = 0.87 \[ a \log {\relax (x )} - a \log {\left (a x + 1 \right )} + a \operatorname {acoth}{\left (a x \right )} - \frac {\operatorname {acoth}{\left (a x \right )}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(a*x)/x**2,x)

[Out]

a*log(x) - a*log(a*x + 1) + a*acoth(a*x) - acoth(a*x)/x

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